Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}9x+6y &= 1 \\ 8x+9y &= 7\end{align*}$
Answer: Begin by moving the $x$ -term in the second equation to the right side of the equation. $9y = -8x+7$ Divide both sides by $9$ to isolate $y$ $y = {-\dfrac{8}{9}x + \dfrac{7}{9}}$ Substitute this expression for $y$ in the first equation. $9x+6({-\dfrac{8}{9}x + \dfrac{7}{9}}) = 1$ $9x - \dfrac{16}{3}x + \dfrac{14}{3} = 1$ Simplify by combining terms, then solve for $x$ $\dfrac{11}{3}x + \dfrac{14}{3} = 1$ $\dfrac{11}{3}x = -\dfrac{11}{3}$ $x = -1$ Substitute $-1$ for $x$ back into the top equation. $9( -1)+6y = 1$ $-9+6y = 1$ $6y = 10$ $y = \dfrac{5}{3}$ The solution is $\enspace x = -1, \enspace y = \dfrac{5}{3}$.